BdMO Online Forum

We'll find general solution for $N=\{1,2,3.......n\}$
Let $\{a,b,c\}$ is a non-$4$-splittable subset. W.L.O.G. $n\geq a>b>c \geq 1$ . Now if $n\geq (a+b-c)=k$ then it will be 4 splittable (as, $k=(a+b-c)>a$). So, $a+b-c>n$______(Condition_1)
Note that $a>(b+c-a)=t$ not belongs to $\{a,b,c\}$; otherwise 2 of them will be same. So $t=(b+c-a)<1$ otherwise it will become 4-splittable. So, $(a-b-c)\geq 0$_____________(Condition_2)
Now $a>(a+c-b)>c$ implies $b=a+c-b$. Otherwise, the 4th element will be $(a+c-b)$ to make it 4-splittable.
So, $a+c=2b$_________(Condition_3)
Now, if $a-b-c=0$ then, $b=2c$ and then $a=3c$...........(s_1)
Otherwise $a-b-c=b-2c>0$ and $b>(b-2c)$. So, $b-2c=c$. (otherwise the 4th element will be $b-2c$ and it will become 4-splitable)
So, here $b=3c,a=5c$...........(s_2)
For (s_1) type solutions; note that $c>\frac{n}{4}$ (Condition_1) and $3c=a\leq n$. So, $c\leq \frac{n}{3}$

For (s_2) type solutions; note that $c>\frac{n}{7}$ (Condition_1) and $5c=a\leq n$. So, $c\leq \frac{n}{5}$

So total number of all non-4-splittable sets $\left \lfloor \frac{n}{3} \right \rfloor - \left \lfloor \frac{n}{4} \right \rfloor + \left \lfloor \frac{n}{5} \right \rfloor- \left \lfloor \frac{n}{7} \right \rfloor$

P.S: I have to say "Cool problem"

As, $AB||PT$, $P$ is the midpoint of minor arc $BC$ and using cyclic property : $BP=CP=AT$ and so $AP||CT$ (Note that $ABPT,APCT$ cyclic trapezium and the two non-parallel sides of a cyclic trapezium are equal). Similarly in trapezium $APCT$, $CS=TS$ Now $CS=PR \Leftrightarrow PR=ST \Leftrightarrow \triangle BPR \cong \triangle ATS$
$\Leftrightarrow BR=AS$ in trapezium $BRSA$ $\Leftrightarrow K,L$ are on the perpendicular bisector of $AB$
$\Leftrightarrow KL$ passes through the midpoint of $AB$ (Note that $BP=AT$ in trapezium $ABPT$ and $AP\cap BT=K$)

$f(x) = 0 \forall x \in \mathbb R$ is the only constant solution for the given function.
Let $P(x,y)$ be the assertion \[f(xy+f(x))=xf(y)+f(x)\]
Let $f(x) = f(y) \neq 0$ for some $x,y \in \mathbb R$. Then \[f(xy+f(x))=f(xy+f(y))\] \[\Rightarrow xf(y)+f(x)= yf(x)+f(y)\] \[\Rightarrow x = y\] which implies injectivity.
Now, $P(0,0) \Rightarrow f(f(0)) = f(0)$, or $f(0) = 0$.
And finally $P(x,0) \Rightarrow f(f(x)) = f(x) \Rightarrow f(x) = x \forall x \in \mathbb R$
So, there are two solutions in total $f(x) = 0 \forall x \in \mathbb R$ and $f(x) = x \forall x \in \mathbb R$